Question

examine the continuity of the following function.
f(x)=x-[x].
[x] represents the greatest integer less than or equal to x

Answer 1

the notation is:

Answer 2

yeah.. it is actually like that but i dont have it.
do you know the answer @abb0t

Answer 3

I don't even know your question. Lol

Answer 4

But since you mentioned continuity, I;m guessing this is a limit's question?

Answer 5

yes. exactly.

Answer 6

it is about continuous and discontinuous functions

Answer 7

Well, x itself is continuous. Draw thegraph.

Answer 8

abb0t drew the ceiling function but what you described is the floor function. Most things with ceiling or floor functions are discontinuous.

Answer 9

@abb0t.. but the function has \[\lfloor x \rfloor\]

Answer 10

that's exactly my concern @Valpey..
how should i start?

Answer 11

The graph looks like:

Answer 12

When f(x) jumps from 1 back down to 0 you have a discontinuity. A function is continuous if the limit as you approach any point from the left or the right is the same.

Answer 13

when we talk about limit exist, it has to be RHS=LHS, right?

Answer 14

they must be same as you approach from both sides (left and right)

Answer 15

Consider \[f(3+\epsilon)\] and \[f(3-\epsilon)\] where epsilon is a very small number. Show that as epsilon approaches zero, they don't converge.

Answer 16

then, since [x] represents as i mentioned above, so LHS will not be same as RHS

Answer 17

if we try to solve using limit? @Valpey

Answer 18

Basically, for epsilon positive:\[\lim_{\epsilon \rightarrow 0}f(3+\epsilon)= \lim_{\epsilon \rightarrow 0}(\epsilon) = 0\] Whereas \[\lim_{\epsilon \rightarrow 0}f(3 -\epsilon) = \lim_{\epsilon \rightarrow 0}(1-\epsilon) = 1\]

Answer 19

if we were to choose x=1 and limit approach to 1.. is it possible @Valpey?

Answer 20

if we choose x=1, then f(1)=0

Answer 21

@abb0t the notation is: